n = 4 has Nuclear Fields The fourth realm has 4 dimensions and is from the interaction between the waves from the 1st and 2nd formulae. The first...
sum n = 1 to 4 for sum m = 1 to n gives us 10, 1 + 2 + 3 + 4. a circle of radius r is formed when r \sin \theta interacts with r \cos \theta ....
The formulae are not equations because they are representing objects. Particle nature would occur when a string has several waves that are not...
n = 3 has Electro-magnetic Fields The third realm has 3 dimensions and is from the interaction between 1st formula and 2nd formula and is the...
A rocket is a closed system. If I were standing inside a rocket at the front and threw a heavy ball towards the back, it would produce an initial...
n=2 has Gravity Field The second realm has 2 dimensions and is from the interaction between 1st formula and 2nd formula and is the surface of...
The first formula describes a single structure of 10 waveforms where the frequency is increased by 3. The second formula describes waveforms as a...
Magnets could be used to act against the Earth's magnetic field so the craft is repulsed by the Earth's magnetic field. The magnets would have to be...
I have been looking at describing our universe in terms of wave functions. These 3 functions seem to give a theory of the universe. \displaystyle...
\displaystyle \sum_{n=1}^8 \left( \sum_{m=1}^n \left( \frac {a \left( \sin \left( {3^{( m+(n^2-n)/2)}} \theta / b + k \times (\frac {\pi}{3} )...
\sum_{n=1}^{40} \left( \sum_{m=1}^n \left( \frac {a \sin \left( {3^{( m+(n^2-n)/2)}} \theta / b + l(\frac {2 \pi}{3}) \right) }{3^{(m+(n^2-n)/2)}}...
\sum_{n=1}^4 \left( \sum_{m=1}^n \left( \frac {a \cos \left( {3^{( m+(n^2-n)/2)}} \theta / b \right) }{3^{(m+(n^2-n)/2)}} \right) \right)...
\displaystyle\sum_{n=1}^4 \left( \sum_{m=1}^n \left( \frac{a \cos\left( {3^{( m+(n^2-n)/2)}} \theta / b \right) }{3^{(m+(n^2-n)/2)}} \right) \right)...
\sum_{n=1}^40 \left( \sum_{m=1}^n \left( \frac {a \sin \left( {3^{( m+(n^2-n)/2)}} \theta / b + l(\frac {2 \pi}{3}) \right) }{3^{(m+(n^2-n)/2)}}...
Your definition of the centre of gravity results in a line through the body. I can balance an object on a point, so there would be no torque from...
You are right for a uniform gravitational pull, a uniform gravitational field, but this occurrance does not happen in any large volume of space.
If you had a large asteroid like 2 balls joined by a thin bar, then the ball closer to a larger body would experience a slightly greater force from...
I have tried to post this <img src="/cgi-bin/mathtex.cgi?\displaystyle\sum_{n=1}^4 \left( \sum_{m=1}^n \left( \frac{a \cos\left( {3^{( m+(n^2-n)/2)}}...
I had always thought of mass to do with inertia, so the centre of mass would be the point around which an object would spin if a force was applied to...
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